Linear Programming Lecture Notes - Penn State Personal Web Server

Figure 1.5. A Level Curve Plot with Gradient Vector: We’ve scaled the gradient
vector in this case to make the picture understandable. Note that the gradient
is perpendicular to the level set curve at the point (1, 1), where the gradient was
evaluated. You can also note that the gradient is pointing in the direction of steepest
ascent of z(x, y).
Proof. As stated, let c(t) be a curve in S. Then c : R → Rn and z(c(t)) = k for all
t ∈ R. Let v be the tangent vector to c at t = 0; that is:
dc(t)
(1.20) = v
dt
t=0
Differentiating z(c(t)) with respect to t using the chain rule and evaluating at t = 0 yields:
(1.21)
d
dt z(c(t))
= ∇z(c(0)) · v = ∇z(x0) · v = 0
t=0
Thus ∇z(x0) is perpendicular to v and thus normal to the set S as required.
Remark 1.19. There’s a simpler proof of this theorem in the case of a mapping z : R 2 →
R. For any such function z(x, y), we know that a level set is an implicitly defined curve given
by the expression
z(x, y) = k
where k ∈ R. We can compute the slope of any tangent line to this curve at some point
(x0, y0) with implicit differentiation. We have:
d
d
z(x, y) = k
dx
dx
yields:
∂z ∂z dy
+
∂x ∂y dx
= 0
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